What is the total angular momentum for non "bound" states? My question has to do with the computation of the parity of a state. For example, if we want to compute the intrinsic parity of the $\pi_0$ meson, we do the following : $P(u\overline{u}) = P(u)P(\overline{u})(-1)^l = -1$. Since $P(u) = -P(\overline{u})$ and $l = 0$ for the $\pi_0$ meson.
Say now we have the decay $\pi_0 \rightarrow \nu_e + \overline{\nu}_e$. If I want to compute the intrinsic parity of the final state, do I have to take into account an eventual $l$ due to the "joint" angular momentum of the system $\nu_e + \overline{\nu}_e$ ?
In other words, do I have $P(\nu_e+\overline{\nu}_e)= P(\nu_e)P(\overline{\nu_e})(-1)^l$ or $P(\nu_e)P(\overline{\nu_e})$.
To extend, what about in general for a decay $a \rightarrow b+c$. Can $b$ and $c$ always have a relative angular momentum $l$? If not, under which condition can they have one? (I know that in $\rho_0 \rightarrow \pi^+ + \pi^-$ we have to take into account a total $l$).
 A: Yes, orbital angular momentum is included when calculating the parity of a final state.
Note that decays to states with large $\ell$ are suppressed compared to states with small $\ell$. A hand-waving way to think about this is to imagine that the initial and final wavefunctions must overlap and recall the radial hydrogen wavefunctions, which go like $(r/a)^\ell$ near the origin; the length scale $a$ for a decay is set by the wavelengths of the particles in the final states.
I vaguely recall that "allowed," "first-forbidden," "second-forbidden," etc. beta-decays are grouped by the orbital angular momentum which must be carried by the decay products.
Your proposed $\pi\to\nu\bar\nu$ is a thorny example: the pion hasn't any spin, but a $\nu\bar\nu$ traveling back-to-back must have total spin $\hbar$.  If that decay happens (current upper limit on the branching ratio is $10^{-6}$) it must have $\ell=1$ to conserve angular momentum. But parity isn't conserved in weak decays anyway.  You might have more luck thinking about $\pi\to\gamma\gamma$, which is the dominant decay mode for the $\pi^0$.
